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To test the effect of alcohol in increasing the reaction time to respond to a given stimulus, the reaction times of seven people were measured. After consuming 3 ounces of \(40 \%\) alcohol, the reaction time for each of the seven people was measured again. Do the following data indicate that the mean reaction time after consuming alcohol was greater than the mean reaction time before consuming alcohol? Use \(\alpha=.05 .\) Person 1 2 3 4 5 6 \begin{tabular}{l} I \\ \hline \end{tabular} \begin{tabular}{llllllll} \hline Before & 4 & 5 & 5 & 4 & 3 & 6 & 2 \\ After & 7 & 8 & 3 & 5 & 4 & 5 & 5 \end{tabular} .

Short Answer

Expert verified
Answer: No, there isn't enough evidence to support the alternative hypothesis at the 0.05 significance level, as the calculated t-statistic (1.63) is less than the critical value (1.943).

Step by step solution

01

Calculate the differences

First, we need to find the difference between the reaction times before and after consuming alcohol for each person: \(d_1 = 7 - 4 = 3\)\ \(d_2 = 8 - 5 = 3\)\ \(d_3 = 3 - 5 = -2\)\ \(d_4 = 5 - 4 = 1\)\ \(d_5 = 4 - 3 = 1\)\ \(d_6 = 5 - 6 = -1\)\ \(d_7 = 5 - 2 = 3\)
02

Compute the mean and standard deviation of the differences

Next, we need to compute the mean (\(\bar{d}\)) and the standard deviation (\(s_d\)) of these differences. \(d = \{3, 3, -2, 1, 1, -1, 3\}\)\ \(\bar{d} = \frac{3+3-2+1+1-1+3}{7} = \frac{8}{7} \approx 1.14\)\ \(s_d = \sqrt{\frac{(3-1.14)^2+(3-1.14)^2+(-2-1.14)^2+(1-1.14)^2+(1-1.14)^2+(-1-1.14)^2+(3-1.14)^2}{6}} \approx 1.86\)
03

Conduct a paired t-test

Since we have the mean and standard deviation of the differences, we can now compute the t-statistic: \(t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} = \frac{1.14}{1.86/\sqrt{7}} \approx 1.63\)
04

Determine the critical value

We need to compare the calculated t-statistic with the critical value at a significance level of \(\alpha = 0.05\). Since this is a one-tailed test, we look up the critical value in a t-distribution table for \(6\) degrees of freedom (\(n-1\)): \(t_{critical} \approx 1.943\)
05

Compare the t-statistic with the critical value

We compare our t-statistic to the critical value: \(t = 1.63 < t_{critical} = 1.943\)
06

Make a conclusion

Since the calculated t-statistic is less than the critical value, we fail to reject the null hypothesis. Thus, there isn't enough evidence to conclude that the mean reaction time after consuming alcohol is greater than the mean reaction time before consuming alcohol at the \(0.05\) significance level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paired t-test
In statistics, the paired t-test is used when you want to compare two related samples. It's most helpful when you have two measurements taken on the same group before and after a treatment or an event. In the example of the reaction time study, we measure the reaction times of individuals before and after they consumed alcohol.
A paired t-test helps determine whether there's a statistically significant difference between the two sets of data. We calculate the difference for each pair (or individual) rather than looking at the raw scores directly. Then we find the mean and standard deviation of these differences.
Key Steps:
  • Calculate the differences for each pair.
  • Determine the mean and standard deviation of these differences.
  • Use these to calculate the t-statistic.
The test then involves comparing this calculated t-statistic to a critical value from the t-distribution table. If the t-statistic is greater than or equal to this critical value, we can say the data supports the hypothesis that the treatment had an effect.
Reaction Time Study
A reaction time study aims to determine how quickly someone responds to a stimulus. In this particular study, researchers were investigating the effect of alcohol on reaction times. They measured how long it takes for participants to react to a signal, both before and after consuming alcohol.
The key idea is to understand if alcohol increases or decreases these reaction times. By conducting such a study, researchers can evaluate the effect that alcohol might have on cognitive functions, which directly relates to activities such as driving.
Think of it this way: if someone has slower reaction times after consuming alcohol, this information could be critical in forming policies, like drink-driving laws or advice on safe alcohol consumption in different scenarios. This study's setup shows how scientific inquiries can have real-world impacts on understanding and improving public safety.
Significance Level
The significance level, denoted by alpha ( extalpha ), is a crucial concept in hypothesis testing. It's the probability of rejecting the null hypothesis when it is actually true, also known as the chance of making a Type I error. In simpler terms, it's how much risk the researcher is willing to take in concluding that there is an effect when there isn't one.
In our original exercise, the significance level is set at 0.05. This translates to a 5% risk of concluding that alcohol does increase reaction times when, in fact, it doesn't. A 0.05 level is a common threshold in scientific studies, striking a balance between being too conservative and too liberal in testing claims.
To decide if the results are statistically significant, compare the t-statistic to the critical value from the t-distribution. If it's smaller, we cannot confidently reject the null hypothesis, suggesting our data doesn't provide enough evidence to prove the effect is real at the specified significance level.

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Most popular questions from this chapter

Two independent random samples of sizes \(n_{1}=4\) and \(n_{2}=5\) are selected from each of two normal populations: \begin{tabular}{l|ccccc} Population 1 & 12 & 3 & 8 & 5 & \\ \hline Population 2 & 14 & 7 & 7 & 9 & 6 \end{tabular} a. Calculate \(s^{2},\) the pooled estimator of \(\sigma^{2}\). b. Find a \(90 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right),\) the difference between the two population means. c. Test \(H_{0}:\left(\mu_{1}-\mu_{2}\right)=0\) against \(H_{\mathrm{a}}:\left(\mu_{1}-\mu_{2}\right)<0\) for \(\alpha=.05 .\) State your conclusions.

Instrument Precision A precision instrument is guaranteed to read accurately to within 2 units. A sample of four instrument readings on the same object yielded the measurements \(353,351,351,\) and \(355 .\) Test the null hypothesis that \(\sigma=.7\) against the alternative \(\sigma>.7\). Use \(\alpha=.05\)

Refer to Exercise 10.91 . Suppose that the word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows: \begin{tabular}{ccc} Person & Stimulus 1 & Stimulus 2 \\ \hline 1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 2 & 1 \\ 5 & 1 & 2 \\ 6 & 2 & 3 \\ 7 & 3 & 3 \\ 8 & 2 & 3 \end{tabular} Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)

An experiment was conducted to compare the mean reaction times to two types of traffic signs: prohibitive (No Left Turn) and permissive (Left Turn Only). Ten drivers were included in the experiment. Each driver was presented with 40 traffic signs, 20 prohibitive and 20 permissive in random order. The mean time to reaction and the number of correct actions were recorded for each driver. The mean reaction times (in milliseconds) to the 20 prohibitive and 20 permissive traffic signs are shown here for each of the 10 drivers: \begin{tabular}{ll} Driver & Prohibitive & Permissive \\ \hline \end{tabular} \(\begin{array}{rrr}1 & 824 & 702 \\ 2 & 866 & 725 \\ 3 & 841 & 744 \\ 4 & 770 & 663 \\ 5 & 829 & 792 \\ 6 & 764 & 708 \\ 7 & 857 & 747 \\ 8 & 831 & 685 \\ 9 & 846 & 742 \\ 10 & 759 & 610\end{array}\) a. Explain why this is a paired-difference experiment and give reasons why the pairing should be useful in increasing information on the difference between the mean reaction times to prohibitive and permissive traffic signs. b. Do the data present sufficient evidence to indicate a difference in mean reaction times to prohibitive and permissive traffic signs? Use the \(p\) -value approach. c. Find a \(95 \%\) confidence interval for the difference in mean reaction times to prohibitive and permissive traffic signs.

Here are the prices per ounce of \(n=13\) different brands of individually wrapped cheese slices: \(\begin{array}{lllll}29.0 & 24.1 & 23.7 & 19.6 & 27.5\end{array}\) 28.7 23.9 \(\begin{array}{lll}21.6 & 25.9 & 27.4\end{array}\) Construct a \(95 \%\) confidence interval estimate of the underlying average price per ounce of individually wrapped cheese slices.

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